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fiasKO [112]
3 years ago
9

Using the graph predict how many paper clipsa 7.5 v battery would pick up for both the 25 coil electromagnet and the 50 coil ele

ctromagnet

Physics
1 answer:
enyata [817]3 years ago
3 0

The graph between the strength of the magnet(number of paper clips picked) and battery is approximately a straight line.

For 25 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 5 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.

For 50 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 15 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.

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How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?
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The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

  • diameter of the steel, d = 4 mm
  • the radius of the wire, r = 2mm = 0.002 m
  • original length of the wire, L₁
  • final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
  • extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
  • the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}

F = \frac{200 \times 10^9\  \times\  1.257\times 10^{-5}\  \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915

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2 years ago
Someone help me please
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